Question

Some of the areas of two squares is 640 metre square if the difference of their perimeter is 64 find the sides of two square

Answer 1

Hi there!

It is given that sum* of the areas of two square is 640m².

Let side of first square be x and side of second square be y

Area of square = (s)²

x² + y² =640 -----------(1)

If the difference of their perimeter is 64

Perimeter of a square = 4.side

4x -4y =64

= x-y=16

x=16+y -----------(2)

Put this value of x in equation (1)

(16+y)² + y²=640

256+y²+ 32y+y²=640

2y² +32y =384

y²+16y=192

y²+ 24y-8y -192=0

y(y+24) -8(y+24)=0

(y-8)=0

(y+24)=0

y= 8

Ignoring negative value i.e -24

Put y= 8 in equation (2)

x=16+y

x=16+8

and x=24

Thus, the side of first square 24cm and second 8cm.

Thankyou:)

Answer 2

Answer:-

8 cm and 24 cm.

Given :-

Sum of the areas of two square is 640 m².

Difference of their perimeter is 64 cm.

To find :-

The sides of two square.

Solution :-

Let the side of first square be x and second square be y.

Area of square is given by :-

$\huge \boxed{A = (Side) ^2}$

A/Q

$x^2 + y^2 = 640----1$

$4x - 4y = 64$

→$4 ( x-y) = 64$

→$x-y = \dfrac{64}{4}$

→$x - y = 16$

→$x = 16 + y$

Put the value of x in eq. 1

→$(16+y) ^2 +y^2 = 640$

→$256 + y^2 + 32y +y^2 = 640$

→$2y^2 +32 y +256 -640 = 0$

→$2y^2 + 32y - 384= 0$

→$2 (y^2 +16y -192 ) = 0$

→$y^2 +16y -192 = 0\times 2$

→$y^2 +24y -8y -192 =0$

→$y(y+24)-8(y+24)=0$

→$(y-8) (y+24) = 0$

→$y -8 = 0 , y +24 = 0$

→$y = 8 , y = -24$

We have to neglect the negative value.

x = 16 + y

x = 16 + 8

x = 24

hence,

The sides of square will be 24 and 8 cm respectively.